Vedanta Excel in Opt. Mathematics Book 7 Final (2078)

Transformation

Example 2. If P(4, 1), Q(5, –1), and R(6, 4) are the vertices of 'PQR. Find the
coordinates of its image under the rotation through +90° about
origin. Show both object and image in the same graph.

Solution: Here, P(4, 1), Q(5, –1), and R(6, 4) are the vertices of 'PQR.

Under the rotation of +90° about origin.

P(x, y) P'(–y, x)

Now, P(4, 1) P'(–1, 4)
Q(5, –1) Q'(1, 5)

R(6, 4) R(–4, 6)

? 'PQR 'P'Q'R'

Y

R'(-4, 6) 7 Q'(1, 5) R(6, 4)

6

5

P'(-1, 4) 4
3

X' 2 P(4, 1) X
1

-7 -6 -5 -4 -3 -2 -1-O1 1 2 3 45678

-2 Q(5, -1)

Y'

Example 3. Let P(5, 2), Q(3, 1), and R (2, –4) be the vertices of 'PQR. 'PQR is
Solution: rotated through +90° about origin and image 'P'Q'R' is formed.
Draw 'PQR and 'P'Q'R' on the same graph.

Rotating the 'PQR through +90° about origin, we have,

P(x, y) P'(–y, x)

Now, P(5, 2) P'(–2, 5)

Q(3, 1) Q'(–1, 3)

R(2, –4) R'(4, 2)

Again rotating 'P'Q'R' through +180° about origin, we get,

P(x, y) P'(–x, –y)

Now, P'(–2, 5) P"(2, –5)

Q'(–1, 3) Q"(1, –3)

R'(4, 2) R"(–4, –2)

? 'PQR 'P'Q'R' 'P"Q"R"

'PPQR and 'P'Q'R' are plotted in the graph.

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Transformation

Y

7

P' 6
5

4

Q' 3 R' P
2 Q

1
X' -7 -6 -5 -4 -3 -2 -1-O1 1 2 3 4 5 6 7 8 X

-2

-3

-4 R

-5

Y'

Exercise 6.2

Short Questions :
1. Find the image of P(a, b) when it is rotated:

(a) through 90° in anticlockwise about origin.
(b) through 90° in clockwise about origin.
(c) through 180° about origin.
2. Draw the images of the following figures through the angles mentioned taking
O as the center of rotation:

(a) A (Through –90°) (b) A (Through +90°)

B

OO

(c) B (d) P (through –90°)

(Through +90°)

A C Q R
O O

104 vedanta Excel in Additional Mathematics - Book 7

(e) M (through –90°) (f) Transformation
R
O P P
N (through –180°) (through +180°)

(g) A Q
O

B

O
C

3. Find the images of the following points when they are rotated through 90°.

(I) Clockwise (II) Anti-clockwise about origin

(a) P(a, b) (b) A(4, 5) (c) B(–6, 6)

(d) C(– 4, – 5) (e) D(5, – 6) (f) E(4, 0)

Long Questions :

4. (a) A(3, 4), B(2, 1), and C(7, 2) are the vertices of 'ABC. Find the coordinates
of its image vertices under the rotation through +90° about the origin.
Present both 'ABC and its image 'A'B'C' on the same graph.

(b) P(3, 4), Q(6, 10), and R(8, 3) are the vertices of 'PQR. Find the coordinates
of its image vertices under the rotation through 90° in anti-clockwise
direction. Plot both triangles on the same graph.

(c) M(2, 1), N(4, 3), and R(1, 4) are the vertices of 'MNP. Find the coordinates
of image vertices of the triangle when it is rotated through 90° in
anti-clockwise about the origin.

5. (a) A(1, 2), B(6, 2), and C(5, 6) are the vertices of 'ABC. The 'ABC is rotated
through an angle of –90° about the origin to get its image 'A'B'C'. Defermine
the coordinates of A', B' and C' and plot 'ABC and 'A'B'C' in the same
graph.

(b) M(2, 5), N(3, 1), and P(5, 6) are the vertices of 'MNP. Rotate 'MNP through
90° in clockwise direction about the origin present both of the triangles on
the same graph.

vedanta Excel in Additional Mathematics - Book 7 105

Transformation

(c) U(2, 4) and V(–2, –4) are the two points of line segment UV. Rotate the line
segment UV through 90° in clockwise direction about the origin. Plot both
of the line segments on the same graph.

6. (a) P(6, 0), Q(8, 3) and R(2, 6) are the vertices of 'PQR. Find the coordinates of
its image under the rotation through 180° about origin. Present the object
and image in the same graph paper.

(b) A(2, 1), B(3, 6) and C(0, 4) are the vertices of 'ABC. Rotate the 'ABC
through 180° in anticlockwise about origin. State the coordinate of the
image vertices present both of the triangles on the same graph.

(c) A unit square with vertices O(0, 0), A(1, 0), B(1, 1) and C(0, 1) is plotted on
a graph. Rotate the square through 180° in clockwise about the origin state
the coordinates of image vertices of the square. Also present the image of
the unit square on the same graph. (Hints: take 50 small divisions - 1 unit)

Project Work

7. Prepare a report on rotation including the following points :
(a) Introduction of rotation
(b) Examples of rotations
(c) Daily use of rotations

1. (a) P'(–b, a) (b) P'(b, –a) (c) P'(–a, –b)

3. (I) (a) P'(b, –a) (b) A'(4, –5) (c) B'(6, 6) (d) O'(–5, 4)

(e) D'(–6, –5) (f) E'(0, –4) (II) (a) P'(–b, a) (b) A'(–5, 4)

(c) B'(–6, –6) (d) C'(5, –4) (e) D'(6, 5) (f) E(0, 4)

4. (a) A'(–4, 3), B'(–1, 2), C'(–2, 7) (b) P'(–4, 3), Q'(–10, 6), R'(–3, 8)

(c) M'(–1, 2), N'(–3, 4), R'(–4, 1) 5.(a) A'(2, –1), B'(2, –6), C'(6, –5)

(b) M'(5 –2), N'(1, –3), P'(6, –5) (c) U'(4, –2), V'(–4, 2)

6. (a) P'(–6, 0), Q'(–8, –3), R'(–2, –6) (b) A'(–2, –1), B'(–3, –6), C'(0, –4)

(c) O'(0, 0), A'(–1, 0), B'(–1, –1), C'(0, –1)

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7

7.0 Introduction

Statistics is a branch of applied mathematics which deals with numerical data. It
generally deals with data collection, tabulation of data, analysis of data, and drawing
the required conclusion for the purpose of the studies.

Nowadays, there is hardly any field of human activities where statistics is not used.
Statistics is used in every field of studies like economics, research, administrations,
agriculture, industry, engineering, education, etc.

7.1 Frequency

The number of times an observation occurs is called the frequency of the observations.
It is denoted by f.

For example: The marks obtained by 15 students in a unit test in mathematics are
given below. The full marks was 20 and the pass marks was 10.

5, 8, 12, 14, 15, 15, 18, 18, 18, 20, 20, 5, 8, 12, 14

From the data, we can classify the data as follows:

Marks Obtained Tally Marks No. of Students
5 || 2
8 || 2
12 || 2
14 || 2
15 || 2
18 ||| 3

20 || 2
Total 15

From the data collected in a class of 15 students, we have classified the data as
shown above. The marks are called variates. The number of times the variate values
repeat is known as frequency. The total of frequency is the total number of students.

We denote the total frequency or sum of frequency by ∑f or N.

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7.2 Frequency Distribution

A tabular arrangement of data showing the frequency of each variate value is
called a frequency distribution. The table showing data with their corresponding
frequencies is called a frequency distribution table. The frequency distribution can
be divided into three types:

(a) Individual Series.

(b) Discrete Series.

(c) Continuous Series.

(a) Individual Series
Individual series are usually list of datas which are collected in a small scale.
They are also called raw data.

Example : The amount of money brought by 5 students of a school for their tiffin.

Rs. 100, Rs. 120, Rs. 150, Rs. 180, Rs. 200

These kind of data are less in number and need not be arranged in table.

Here, number of datas is denoted by n.

In the above example, n = 5

(b) Discrete Series
The series formed by discrete values are called discrete series. We usually have
variates and frequency in this discrete series.

The marks obtained by 20 students in a unit test are given below:

Marks 7 10 12 14 15 18 20
No of students 3 232523

Here, marks denote the variate values and the number of students denote the
frequencies. Total frequency is denoted by N = ¦f.

(c) Continuous Series

The series which can be represented by a continuous variable is called
continuous series. We usually use this kind of data to insert large scale of datas.
The marks obtained by 100 student in an examination is given below.

Marks 0 - 10 10 - 20 20 - 30 30 - 40 40 - 50
15 25 30 20
No of students 10

In this series, the data lie within groups. Such interval or groups are called
class intervals. The end points of class interval are called limits.

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Example: In case of 0-10 0 is called lower limit. 10 is called upper limit.

The difference between the upper limit and the tower limit of corresponding
class is called class interval. In above example, the class interval is 10.

In continuous class, the upper limit of the class is excluded. So it is also called
exclusive method. In class (0-10), 10 is excluded. That is 10 does not lie in class
(0-10). 10 lies in (10-20).

The middle value of corresponding class is obtained by using the formula

Middle value = upper limit + lower limit
2

In above example

Mid value of class (0-10) = 0 +10 = 5
2
10 + 20
Mid value of class (10-20) = 2 = 15

7.3 Cummulative Frequency (c.f.) Table

The cummulative frequency of a class interval is the sum of frequencies all classes
upto and including that class. A frequency distribution showing the cummulative
frequencies corresponding values of the variable systematically arranged in
increasing (or decreasing) order is known as the cummulative frequency distribution.

In a cummulative frequency table, cummulative frequency corresponding to a class-
interval is the sum of all frequencies upto and including that class interval.

A table which displays the cummulative frequencies are distributive over various
classes is called cummulative frequency table.

Example: Let us consider weights of 50 students given in the table:

Weights (in kg) 20-25 25-30 30-35 35-40 40-45 45-50
8
No of students 2 8 10 12 10

Cummulative Frequency table of above data is given below:

Weights (kg) No. of students (f) Cummulative Frequency (c.f.)
20-25 2 2
25-30 8
30-35 10 2 + 8 = 10
35-40 12 10 + 10 = 20
40-45 10 20 + 12 = 32
45-50 8 32 + 10 = 42
42 + 8 = 50

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Worked out Examples

Example 1. Construct the cummulative frequency table from the following data.

Marks (x) 5 10 15 20 25 30 35
1
Frequency (f) 2 4 6 4 3 2

Solution: Cummulative frequency table of given data is given below:

Marks (x) Frequency (f) Cummulative Frequency (c.f.)
5 2 2
10 4
15 6 2+4=6
20 4 6 + 6 = 12
25 3 12 + 4 = 16
30 2 16 + 3 = 19
19 + 2 = 21
35 1
21 + 1 = 22

Example 2. The marks obtained by 30 students in mathematics are listed below
Solution: construct the cummulative frequency table from the given data.

10 20 24 48 30 32 20 40 10 24

18 35 20 40 48 30 32 20 45 10

24 48 18 30 20 40 24 18 45 30

Cummulative frequency table of given data is given below:

Marks (x) Tally marks No. of students Cummulative
(f) Frequency (c.f.)
10 ||| 3
18 ||| 3 3
20 |||| 5 3+3=6
24 |||| 4 6 + 5 = 11
30 |||| 4 11 + 4 = 15
32 || 2 15 + 4 = 19
35 19 + 2 = 21
40 | 1
45 ||| 21 + 1 = 22
48 || 3
||| 22 + 3 = 25
2
25 + 2 = 27
3
27 + 3 = 30

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Example 3. Prepare a cummulative frequency table from given data:

Marks 0-10 10-20 20-30 30-40 40-50
No. of students 5 15 25 20 5

Solution: The cummulative frequency table of above data is given below:

Marks (x) No. of students (f) Cummulative Frequency (c.f.)
0-10 5 5
10-20 15
20-30 25 5 + 15 = 20
30-40 20 20 + 25 = 45
40-50 5 45 + 20 = 65
65 + 5 = 70
Exercise 7.1

1. Construct a cummulative frequency table from the following data:

(a) Marks obtained 10 15 20 25 30 35 40
2
No. of students 2 4 6 8 5 3

(b) x 20 30 40 60 70 80 90
f 2 5 10 7 6 3 2

(c) Marks 0-10 10-20 20-30 30-40 40-50
No. of students 6 12 15 12 5

(d) Age (years) 0-10 10-20 20-30 30-40 40-50 50-60
No. of people 10 15 25 15 10 52

2. (a) Ages of 30 students of class VII are given below. Prepare a cummulative
frequency table from the given data:

12, 16, 13, 16, 13, 12, 16, 15, 13, 14,

14, 15, 15, 14, 14, 16, 14, 12, 14, 15,

11, 16, 12, 13, 11, 13, 16, 11, 14, 15

(b) The weight (in kgs) of 25 students are given below. Prepare a cummulative
frequency table:

25, 30, 35, 25, 26, 40, 25, 28, 28, 35,

28, 40, 45, 28, 40, 40, 25, 40, 45, 48,

30, 28, 48, 30, 48

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3. (a) Given data is marks of 40 students of class VII. Construct a frequency
distribution table taking the class interval of 10. Also prepare a cummulative
frequency table:
35, 40, 45, 50, 70, 80, 90,, 55, 65, 40,
40, 45, 60, 75, 80, 54, 55, 58, 60, 75,
90, 70, 40, 55, 75, 85, 90, 45, 60, 88,
45, 60, 75, 85, 55, 94, 75, 50, 45, 68

(b) Given data is weights (kg) of 30 students of class VII. Make a frequency
table taking class interval 5. Also prepare a cummulative frequency table:
45, 44, 33, 35, 30, 50, 44, 60, 52, 60,
62, 64, 44, 55, 35, 45, 44, 50, 32, 35,
42, 46, 37, 40, 44, 47, 62, 60, 58, 50

7.4 Arithmetic Mean

Average of a set of data is very important. It daily life we use 'average' to represent
arithmetic mean. Arithmetic mean of set of data can represent whole set of data. It
usually represents the central data. The percentage of any student is the average
marks of all the marks they get in the exam.

The arithmetic mean of the given data is calculated by adding all the data together
and dividing them by the total number of data.

Mathematically,

Mean = Sum of the data
X Total number of data
= ¦nX,

where X is the variate values.

¦X is the sum of all variates

n is total number of data

Indiscrete Series.

In case of discrete series or ungrouped repeated data, we find mean by using formula.

Mean (x) = ¦fx
N

Where, x is the variate

f is the frequency

fx is the product of f and x

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∑fx is the sum of fx.
N is the total number of frequency.

Worked Out Examples

Example 1. Find the mean of 10, 20, 30, 40, 50
Solution:
Here, the data are 10, 20, 30, 40, 50

number of data (n) = 5

∑x = 10 + 20 + 30 + 40 + 50

= 150 = ¦x
Now, mean (x) n

= 150
5

= 30

Example 2. The average of, 10, 20, 30, p, 50, and 60 is 35. Find the value of p.
Solution:
Here, the data are 10, 20, 30, p, 50, 60

Number of data (n) = 5

∑x = 10 + 20 + 30 + p + 50 + 60

= 170 + p

Now, we know,

mean (X) = 35

So, mean (X ) = ¦X
n
170 + p
or, 35 = 6

or, 210 = 170 + p

or, p = 40

? The value of p is 40.

Example 3. Calculate the mean from the data given below:

Marks 10 20 30 40 50

No. of students 4 6 10 6 4

Solution: Calculation of Mean,

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Statistics

Marks (x) No. of students (f) fx
40
10 4 120
300
20 6 240
200
30 10 ¦fx = 900

40 6

50 4

N = 30

Now, Mean (x) = ¦fx = 900 = 30
N 30

Hence, mean marks = 30.

Example 4. Find the value of p from the data given below whose mean is 20:

x 5 10 15 20 25 30 35
f 27p 7 644

Solution: Calculation of missing frequency: fx
10
xf 70
52 15p
10 7 140
15 p 150
20 7 120
25 6 140
30 4 ¦fx = 630 + 15p
35 4

¦f = 30 + p

By using formula,

Mean (x) = ¦fx
N
630 + 15p
So, 20 = 30 + p

or, 600 + 20p = 630 + 15p

or, 30 = 5p

? p=6

Hence, the missing frequency is 6.

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Exercise 7.2

1. Find the mean from the data given below:
(a) 35, 40, 45, 50, 60
(b) 10, 11, 14, 16, 17, 19, 13, 20, 25, 21

2. (a) If the mean of 5, 12, 8, 10, and p is 8, find the value of p.

(b) If the mean of 15, 21, 25, and p is 20, find the value of p.
Long Questions :
3. (a) From the data given below calculate the mean:

Marks 20 40 60 80 100
3
No of students 6 9 4 3
500
(b) From the data given below, find the mean: 10

Wages (in Rs.) 100 200 300 400

No. of workers 10 15 10 15

(c) x 5 10 15 20 25 30
f 2 5 10 7 4 2

(d) x 58 60 62 64 66 68

f 12 14 20 13 8 5

(e) Marks 40 50 55 62 75 80

No. of students 4 6 10 8 5 2

4. (a) The mean of the data given below is 21. Find the value of p.

x 10 15 20 25 35
f 56p65

(b) The mean of the data given below is 34. Find the missing frequency.

x 10 20 30 40 50 60
f 24q432

Project Work

5. List the marks obtained by the students of your class in mathematics in the
class 6. Then find the average of the marks.

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Statistics

1. (a) 46 (b) 16.6 2.(a) 5 (b) 19
3. (a) 50.4 marks (b) Rs. 300
4. (a) 3 (b) 5 (c) 17 (d) 62.17 (e) 58.31 marks

7.5 Median

The data that divides the given set of data into two equal halves is known as median.

It is denoted by Md. To calculate median, we need to arrange the data either in
ascending or descending order.

(a) For Individual Series

Median (Md) = value of n+1 th
2
item

where, n = total number of items

(b) For Discrete Series
Steps for calculation of Median

(i) First arrange the variables in ascending order.

(ii) Write commulative frequency in c.f. column.

(iii) Find N+ 1 , where N = total frequency.
2

(iv) Then observe the c.f. just greater than or equal to N+ 1 at c.f. column,
2
N+ 1
then the corresponding value of 2 in variate column is the required

median. If any c.f. is just greater than N+ 1 of c.f. column, then the
2

value corresponding to that c.f is the required median.

Worked out Examples

Example 1. Find the median from the set of data given below:
Solution: 30, 70, 90, 100, 150
Here, the given items are 30, 70, 90, 100, 150

116 vedanta Excel in Additional Mathematics - Book 7

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which are already in ascending order.

So, number of items (n) = 5

Now, median (Md) = value of (n + 1)th item
2

= value of (5 + 1)th item
2

= value of 6 th
2
item

= value of 3rd item

? Median (Md) = 90
Example 2. Find the median from the given data : 22, 18, 14, 12, 24, 26, 32, 34
Solution: Here, the given data are arranged in ascending order

12, 14, 18, 22, 24, 26, 32, 34

Now, number of items (n) = 8

So, Median (Md) = value of (8 + 1)th item
2

= value of 9 th item
2

= value of 4.5th item

Now 4. 5th item is not present

So, 4.5th item is the mean of 4th item and 5th item

So, median (Md) = value of 4th item + value of 5th item
2

= 22 + 24
2

= 46
2

= 23

? Median (Md) = 23

Example 3. 16, 18, 20, x, 24, 26, 28 are in ascending order. If the median of given
Solution: data is 22, find the value of x.

Here, the given items are 16, 18, 20, x, 24, 26, 28

the number of items (n) = 7

? Also, median (Md) = 22

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We know, = value of (n + 1)th item
Median (Md) 2

= value of (7 + 1)th item
2

= value of 4th item

=x

But, by the question,

Md = 22
? x = 22

Example 4. 5, p – 12, p + 11 and 42 are in ascending order. If the median of the
data is 19, find the value of p.

Solution: Here, the given items are 5, p – 12, p + 11 and 42

number of items (n) = 4

? Now, median (Md) = value of (n + 1)th item
2
(4 + 1)th
= value of 2 item

= value of 2.5th item

= value of (2nd item + 3rd item)
2
p – 12 + p + 11
= 2

= 2p – 1
2

But, by the question,

Md = 19

So, 2p – 1 = 19
2

or, 2p – 1 = 38

or, 2p = 39

? p = 19.5

Example 5. Find the median from given data:

x 5 10 15 20 25 30 35 40
f 4 6 10 15 10 8 6 4

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Solution: To find median, given data can be written in c.f. table.

x f Cummulative frequency (c.f.)
5 4 4
10 6
15 10 4 + 6 = 10
20 15 10 + 10 = 20
25 10 20 + 15 = 35
30 8 35 + 10 = 45
35 6 45 + 8 = 53
40 4 53 + 6 = 59
Total N = 63 59 + 4 = 63

Here, N = 63

we have, using formula,

Median (Md) = N+1 th
2
item

= 63 + 1 th
2
item

= 32nd item

c.f. just greater 32 is 35 whose corresponding value is 20.

? Median (Md) = 20

Exercise 7.3

1. Find the median from the data given below:

(a) 22, 24, 26, 28, 32 (b) 6, 8, 10, 12, 14, 16, 18

(c) 100, 200, 300, 400, 500, 600, 700

2. Find the median of the given data:

(a) 28, 21, 24, 20, 27, 25 (b) 40, 44, 48, 60, 56, 72, 65, 53

(c) 100, 400, 600, 800, 1000, 300, 200, 700

3. (a) If the median of the data 13, 14, p, 17, 20 which are arranged in ascending
order is 16, find the value of p.

(b) If the median of the data 400, 350, 300, q, 200, 150, 100 which are arranged
in descending order is 250, find the value of q.

4. (a) The median of the data p – 1, p + 1, 2p + 5, and 3p + 1 are the data in
ascending order is 18, find the value of 'p' and the given numbers.

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(b) If p, p + 4, p + 6, p + 8 are in ascending order and of the median is 20, find
the value of 'p' and the numbers.

5. Find the median from the following data:

(a) Marks 10 20 30 40 50 60 70

No. of students 2 4 6 10 4 3 1

(b) Weight (kg) 10 15 20 22 24 30 35

No. of children 8 10 16 10 4 6 3

(c) x 100 200 300 400 500 600
f 4 6 10 8 6 2

Project Work

6. Measure the heights of your friends and tabulate them as shown in the given
table.

Name of students Height (feet)
1.
2.
3.
4.
5.

Arrange the heights in ascending order. Find the median height.

1. (a) 26 (b) 12 (c) 400

2. (a) 24.5 (b) 54.5 (c) 500

3. (a) 16 (b) 250 4.(a) p = 10, 9, 11, 25, 31

(b) p = 15; 15, 19, 21, 23 5.(a) 40 (b) 20 (c) 300

7.6 Mode

The variate value which occurs maximum number of times in a distribution is
called mode. In other words, the most frequently repeated value is called the mode.
In a distribution, mode occurs with maximum frequency. It is denoted by MO.

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(a) Mode for Individual Series
In case of individual series, mode can be found by inspection. To find the
mode, count the number of times the various values repeat themselves. Then,
the maximum number of repeated value (i.e., value with maximum frequency)
is the mode. Here, we study mode for individual and discrete series.

Worked Out Examples

Example 1. Find the mode from the following data.
Solution: 15, 10, 14, 15, 18, 20, 15, 14, 13, 10, 15, 18
Calculation of mode

Variate value No. of occurrence (Frequency)
10 2
13 1
14 2
15 4
18 2
20 1

From above table the value 15 is 4 times repeated. 4 is the
maximum frequency whose corresponding value is 15.

? Mode (MO) = 15

(b) Mode for Discrete Series

In this series, mode can be found simply by inspection. The value with the
maximum frequency is the mode.

Example 2. Find the mode from the given table:

Marks 40 50 60 70 80 90 100

No. of students 4 6 12 16 12 4 2

Solution: Here, the maximum frequency is 16 whose corresponding value is 70.

? Mode (MO) = 70

Exercise 7.4

Short Questions :
1. Find the modal value of the following data:

(a) 7, 8, 10, 7, 10, 7, 8, 7, 12, 7, 13, 7, 8

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(b) 2, 4, 6, 8, 6, 8, 7, 8, 8, 10, 12, 6
(c) 100, 200, 300, 300, 400, 500, 300, 100, 400, 500, 300
2. Find the modal value of the following data :

(a) Age (years) 10 12 16 18 20 22
No. of people 4 6 12 8 4 3

(b) x 20 30 40 50 60 70 80
f 2 6 10 15 12 13 6

(c) Size of shoes 3 4 5 6 7 8 9 10
No. of persons 10 15 20 12 8 6 4 2

Long Questions :
3. (a) A boy scored the following marks in 11 tests of different units:

10, 12, 15, 16, 12, 13, 15, 14, 15, 18, 15
(i) What is the modal marks of the boy?
(ii) What is his median marks?
(iii) What is his mean marks?
(b) A Selesman sold the shoes of the following sizes.
2, 4, 6, 4, 6, 8, 4, 6, 8, 7, 6, 2, 5, 8, 6
(i) What is the model size of the shoes?
(ii) Find the median size of the shoes.
(iii) Find the mean size of the shoes.

1. (a) 7 (b) 8 (c) 300
2. (a) 16 (b) 50 (c) 5
3. (a) (i) MO = 15 (ii) Md = 15 (iii) X = 14.09
(ii) Md = 6 (iii) X = 5.47
(b) (i) MO = 6

122 vedanta Excel in Additional Mathematics - Book 7

Statistics

Model Questions

Class : 7 Annual Examination Full marks : 50
Time : 1.5 hours

Attempt all the questions.

Group A [9×2 = 18]

1. Find the values of x and y of x , y = (2, 3).
3 4

2. If A = 1 2 and B = 2 –4 , find (A + B).
4 5 6 2

3. Find the distance between P(6, 7) and Q(–2, 1).

4. Simplify : 32 – 8 + 12 2

5. State Pythagoras theorem with figure.

6. Prove that (1 – sin2A) (1 + cot2A) = cot2A

7. Find the images of P(7, 8) and Q(2, 3) when they are reflected on X-axis.

8. Find the arithmetic mean of given data :

10, 20, 30, 40, 50, 60, 70

9. Find the mode of given data :

6, 8, 12, 8, 12, 14, 16, 18, 12, 15, 12

Group B [8×4 = 32]

10. If R = {(1, 1), (2, 4), (3, 9), (4, 16)} is a relation, show it in the following ways.

(a) Table method 123
(b) Arrow diagram method
(c) Graphical method
(d) Rule method

vedanta Excel in Additional Mathematics - Book 7

Statistics

11. If P = 2 1 , Q = –4 3 , find P – Q and Q – P. Are P – Q and Q – P equal?
4 5 6 4

12. If A(1, 1), B(4, 3), C(2, 1), and D(5, 3), prove that AB = CD.

13. Define cartesian product of two sets. If A = {1, 2} and B = {4, 6, 7}, find

A × B and B × A. A

14. From the given right angled triangle, find the six fundamental 2 2
trigonometric ratios taking reference angle.

15. Prove that 1 + sin + 1 cos = 2sec . C 2B
cos + sin PQR and its
PQR.
16. If P(3, 8), Q(3, 4), and R(6, 7) are the vertices of PQR. Plot
Reflect it on X-axis and find the image vertices of
image on the same graph.

17. Find arithmetic mean from the following data :

Marks 10 20 30 40 50 60 70 80

No. of students 2 4 6 15 12 6 4 3

124 vedanta Excel in Additional Mathematics - Book 7